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\title{Distributed Entanglement Paper Review}
\author{Hu Jingcheng 3180105323@zju.edu.cn}

\begin{document}
\maketitle
\section{Introduction}
In order to understand quantum entanglement, 
a quantitative description of quantum entanglement is certainly need, 
also giving specific guide on how it can and cannot be manipulated.
In this paper, we first give a formal definition of \textbf{concurrence}, 
which quantifies two-qubit entanglement both in pure and mixed state. 
And using this definition, we establish an important inequality which reveals the exisitance of three-way entanglement, 
denoting to be \textbf{resisual entanglement}. We also give an explicit formula of resisual entanglement, 
and prove its invariance under permutation transformation. 
\section{Results}

\subsection{Quantifying Entanglement of Two Qubits: Concurrence}
\subsubsection{Concurrence for Pure State}
Suppose we are given two qubits A,B. 
We first suppose qubits A,B are in pure state $|\Phi\rangle$, 
and then generalize our definition to mixed states.

For state $|\Phi\rangle$, we define "spin-flip" operation:
$$
|\tilde{\Phi}\rangle \triangleq (\sigma_y \otimes \sigma_y) |\Phi^\star \rangle 
$$
where the conjugation is taken in standard basis $\{|00\rangle,|01\rangle, |10\rangle,|11\rangle\}$.
For a single bit, which you may visualize as a vector in Bloch sphere, 
the "spin-flip" operation does the transformation 
equivalent to multiplying -1 to the Bloch vector, 
which is essentially transforming the state to its orthogonal state. 
And this is why it is called "spin-flip" operation.

Then we can define \textbf{concurrence} \cite{concurrence_review} as the quantitative description of entanglement between A and B:
$$
C(\Phi) \triangleq |\langle \Phi | \tilde{\Phi} \rangle |
$$
we have:
$$
\begin{cases}
    C = 0 &\iff \text{not entangled at all} \\
    0<C<1 &\iff \text{partially entangled} \\
    C = 1 &\iff \text{completely entangled} \\ 
\end{cases}
$$
For product state, it is obvious that both single bit state 
will be transformed to its orthogonal state, 
thus $\langle \Phi | \tilde{\Phi} \rangle = 0 \iff C(\Phi) = 0$. 
And for completely entangled state like singlet state, we may easily verify 
that $C = 1$

Now we use algebraic way to give an explicit formula for concurrence, 
which help us understand better and is useful in future deriation.
In standard basis $\{|00\rangle,|01\rangle, |10\rangle,|11\rangle\}$, 
we write $|\Phi\rangle = (a,b,c,d)$, then we have:
\begin{align}\label{eq:1}
|\tilde{\Phi}\rangle = (-d^\star, c^\star, b^\star, -a^\star) \implies C(\Phi) = 2|ad-bc|
\end{align}
Now use this formula, 
we can easily see the relation between concurrence and how two qubits are entangled.

Now we introduce the density operator $\rho_{AB}$ and reduced density operator $\rho_A = tr_B(\rho_{AB})$. 
We now prove a useful formula:
\begin{align}\label{eq:2}
    {C(\Phi)}^2 = 4 \det{\rho_A}, \text{if AB is pure state}
\end{align}
\begin{proof}
AB is in pure state $|\Phi\rangle$. 
Write $|\Phi\rangle$ in standard basis, and repeated indices are summed:
$$
|\Phi\rangle = a_{ij}|ij\rangle, i,j = 0 \text{ or } 1
$$
Then we have:
$$
\rho_{AB} = |\Phi\rangle \langle\Phi| = a_{ij}a^\star_{lm} |ij\rangle\langle lm|\\
$$
$$
\implies \rho_A = tr_B \rho_{AB} = a_{ij}a^\star_{lm} tr_B |ij\rangle\langle lm| = \sum_j a_{ij}a^\star_{lj}|i\rangle\langle l| \\
$$
We can write $\rho_A$ explicitly in matrix form:
$$
\rho_A = 
\begin{bmatrix}
    aa^\star+bb^\star & ac^\star+bd^\star \\ 
    a^\star c + b^\star d & cc^\star + dd^\star
\end{bmatrix}
$$
Then according to \autoref{eq:1}, expand the expression and then we can prove this formula.
\end{proof}

\subsubsection{Concurrence for Mixed State}
Now we give the definition of concurrence when AB is mixed state\cite{concurrence_review}:
$$
C(\rho) = \inf \sum_j p_j C(\Phi_j), \text{ where }\rho = \sum_j p_j |\Phi_j\rangle\langle\Phi_j|
$$
And we give an explicit formula for mixed state concurrence without offering proof here
(if needed, please check the reference\cite{concurrence_mixed_state}).
Now we define:
$$
\tilde\rho = (\sigma_y \otimes \sigma_y) \rho^\star (\sigma_y \otimes \sigma_y)
$$
note that both $\rho$ and $\tilde{\rho}$ are positive operator, 
therefore $\rho\tilde{\rho}$, though not Hermitian, 
has only real and non-negative eigenvalues.
We denote the \textit{square root} of the eigenvalues in decreasing order $\lambda_1,\lambda_2,\lambda_3,\lambda_4$.
Then we have the explicit formula:
\begin{align}\label{eq:3}
    C(\rho) = \max{\{0,\lambda_1-\lambda_2-\lambda_3-\lambda_4\}}
\end{align}
One can easily verify when AB is in pure state, the formula degenerate to \autoref{eq:2}.


\subsection{A Definition of Three-Way Entanglement: Residual Entanglement}
After formalizing our language describing two-qubit entanglement, 
now we are trying to explore how three qubits are entangled and whether there are intrinsic limitations on how we can utilize it.

\subsubsection{Establish Inequality with $C_{AB}$ and $C_{AC}$}\label{sec:2.2.1}
Now we turn to this problem: given a pure state of three qubits A, B, and C,
how is $C_{AB}$ is related to $C_{AC}$ ? We first provide our conclusion:
\begin{align}
    C^2_{AB} + C^2_{AC} \le C^2_{A(BC)}
\end{align}
We first explain that $C_{A(BC)}$ is well-defined. 
And the discussion here is of great importance for understanding future deduction. 

Note that ABC is in pure state, which means:
$$
\rho_{ABC} = |\Phi\rangle \langle\Phi| = a_{ijk}a^\star_{lmn}|ijk\rangle \langle lmn|
$$
\begin{align*}
\implies \rho_{BC} &= tr_A\rho_{ABC}  \\
&= \sum_i a_{ijk}a^\star_{imn}|jk\rangle \langle mn| \\
&= a_{0jk}a^\star_{0mn}|jk\rangle \langle mn| +a_{1jk}a^\star_{1mk}|jk\rangle \langle mn|  = |\Phi_1\rangle \langle\Phi_1| + |\Phi_2\rangle \langle\Phi_2|
\end{align*}
Now we immediately recognize that $\dim \text{ran} (\rho_{BC}) = 2$, 
and $\text{ran} (\rho_{BC}) $ is spanned by $\{|\Phi_1\rangle,|\Phi_2\rangle\}$. 
which also means only at most two of the eigenvalues of the $\rho_{BC}$ is non-zero, 
And it is always convenient to discuss the problem in $\text{ran} (\rho_{BC})$, and in this way 
you can equivalently treat BC as a single qubit in its mixed-state.
Thus you can define two-qubit concurrence $C_{A(BC)}$ properly.
\begin{proof}
Similar to the discussion above, we can deduce 
$$
\rho_{AB} = a_{ij0}a^\star_{lm0}|ij\rangle \langle lm| +a_{ij1}a^\star_{lm1}|ij\rangle \langle lm|=|\Phi_1\rangle \langle\Phi_1| + |\Phi_2\rangle \langle\Phi_2|
$$
$$
\implies \tilde{\rho}_{AB} = (\sigma_y \otimes \sigma_y) \rho^\star_{AB} (\sigma_y \otimes \sigma_y) = |\tilde\Phi_1\rangle \langle\tilde\Phi_1| + |\tilde\Phi_2\rangle \langle\tilde\Phi_2|
$$
Therefore, we know:
$$
\text{ran} (\rho_{AB})=\text{ran} (\tilde\rho_{AB}) = 2 
$$
$$
\implies \text{ran} (\rho_{AB}\tilde\rho_{AB}) \le \max{\{\text{ran} (\rho_{AB}),\text{ran} (\tilde\rho_{AB})\}}=2
$$
Thus $\rho_{AB}\tilde\rho_{AB}$ also has at most two non-zero eigenvalues; 
we might as well denote their square root as $\lambda_1,\lambda_2$, and $\lambda_1\ge\lambda_2 \ge 0$. 
Then using \autoref{eq:3} we have:
\begin{align}\label{eq:4}
C^2_{AB} &= {(\lambda_1 - \lambda_2)}^2\\
& = \lambda_1^2+\lambda_2^2 - 2\lambda_1\lambda_2 \\
& = tr(\rho_{AB}\tilde\rho_{AB})- 2\lambda_1\lambda_2 
\end{align}
\begin{align*}
\implies C^2_{AB} \le tr(\rho_{AB}\tilde\rho_{AB})
\end{align*}
Analogously, for pair AC, we can write:
\begin{align*}
C^2_{BC} \le tr(\rho_{BC}\tilde\rho_{BC})
\end{align*}
Summing up above two inequalities, we obain a very important itermediate inequality:
\begin{align}\label{eq:5}
C^2_{AB}+C^2_{BC} \le tr(\rho_{AB}\tilde\rho_{AB})+tr(\rho_{BC}\tilde\rho_{BC})
\end{align}

Next we try to give a better expression for RHS of \autoref{eq:5}.

First we want to give a explicit formula for $\tilde\rho_{AB}$. 
Note that:
\begin{align*}
\tilde\rho_{AB} &= (\sigma_y \otimes \sigma_y) \rho^\star_{AB} (\sigma_y \otimes \sigma_y)\\
&= (\epsilon \otimes \epsilon) \rho^\star_{AB} (\epsilon \otimes \epsilon)
\end{align*}
where $\epsilon$ is 2-dimentional Levi-Civita tensor, and its matrix form is:
$$
\epsilon = 
\begin{bmatrix}
\epsilon_{00}&\epsilon_{01}\\\epsilon_{10}&\epsilon_{11} 
\end{bmatrix}
= \begin{bmatrix} 0&1\\-1&0 \end{bmatrix}
\implies\epsilon \otimes \epsilon = \epsilon_{ab}\epsilon_{cd}|ac\rangle \langle bd|
$$
Then we can calculate $\rho_{AB}\tilde\rho_{AB}$ explicitly:
\begin{align*}
&\rho_{AB}\tilde\rho_{AB} \\
&= a_{ijk}a^\star_{mnk}|ij\rangle \langle mn|\cdot \epsilon_{ab}\epsilon_{cd}|ac\rangle \langle bd|\cdot a^\star_{m'n'p}a_{i'j'p}|m'n'\rangle \langle i'j'|\cdot \epsilon_{ef}\epsilon_{gh}|eg\rangle \langle fh| \\
&= a_{ijk}a^\star_{mnk}\epsilon_{ab}\epsilon_{cd} a^\star_{m'n'p}a_{i'j'p}\epsilon_{ef}\epsilon_{gh} |ij\rangle \cancelto{\delta_{al}\delta_{cm}\delta_{bm'}\delta_{dn'}\delta_{ei'}\delta_{gj'}}{\big(\langle lm|\cdot |ac\rangle \langle bd|\cdot |m'n'\rangle \langle i'j'|\cdot |eg\rangle\big)} \langle fh| \\
\\
&= a_{ijk}a^\star_{mnk}\epsilon_{mm'}\epsilon_{nn'} a^\star_{m'n'p}a_{i'j'p}\epsilon_{i'f}\epsilon_{j'h} |ij\rangle\langle fh|
\end{align*}
Then we can write down its trace directly:
\begin{align*}
tr(\rho_{AB}\tilde\rho_{AB}) &= \langle ij|\rho_{AB}\tilde\rho_{AB}|ij\rangle\\
&=\langle ij|a_{ijk}a^\star_{mnk}\epsilon_{mm'}\epsilon_{nn'} a^\star_{m'n'p}a_{i'j'p}\epsilon_{i'f}\epsilon_{j'h} |ij\rangle \cancelto{\delta_{fi}\delta_{hj}}{\langle fh||ij\rangle}\\
&=a_{ijk}a^\star_{mnk}\epsilon_{mm'}\epsilon_{nn'}a^\star_{m'n'p}a_{i'j'p}\epsilon_{i'i}\epsilon_{j'j}
\end{align*}
Notice the identity for 2-dimentional Levi-Civita tensor:
$$
\epsilon_{ij}\epsilon_{lm} = \delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}
$$
Plug in the identity, and reorder the factors:
\begin{align*}
tr(\rho_{AB}\tilde\rho_{AB}) &=a_{ijk}a^\star_{mnk}a^\star_{m'n'p}a_{i'j'p} \epsilon_{mm'}\epsilon_{i'i}\cancelto{\delta_{nj'}\delta_{n'j}-\delta_{nj}\delta_{n'j'}}{\epsilon_{nn'}\epsilon_{j'j}}\\
&=a_{ijk}a^\star_{mnk}a^\star_{m'jp}a_{i'np}\epsilon_{mm'}\epsilon_{i'i} +\cancelto{{(\rho_A)}_{im}}{a_{ijk}a^\star_{mjk}}\cancelto{{(\rho_A)}_{i'm'}}{a_{i'n'p}a^\star_{m'n'p}} \epsilon_{ii'} \epsilon_{mm'} \\
&= a_{ijk}a^\star_{ijp}a_{mnp}a^\star_{mnk}-a_{ijk}a^\star_{ink}a_{i'np}a^\star_{i'jp}+2\det \rho_A\\
&= tr\rho^2_C -tr\rho^2_B +2\det \rho_A
\end{align*}
Also notice $tr\rho = 1 \iff \lambda_1+\lambda_2 = 1 \implies tr\rho^2 = \lambda^2_1+\lambda^2_2 = \cancelto{1}{{(tr\rho)}^2} - 2\det\rho$
Then we can rewrite the above identity:
\begin{align}\label{eq:6}
tr(\rho_{AB}\tilde\rho_{AB}) &= 2(\det \rho_A+\det \rho_B-\det \rho_C)
\end{align}
Analogously, we have identity for AC, and combining with \autoref{eq:6} we have:
\begin{align}
RHS = tr(\rho_{AB}\tilde\rho_{AB})+tr(\rho_{AC}\tilde\rho_{AC}) = 4\det \rho_A
\end{align}
Then combining with the discussion of $C_{A(BC)}$ and \autoref{eq:2}, we reach our final results:
\begin{align}\label{eq:11}
    C^2_{AB}+C^2_{BC} \le 4\det \rho_A = C^2_{A(BC)}
\end{align}
\end{proof}

\subsubsection{Using the Inequality to define Resisual Entanglement}
According to \autoref{eq:4} and the inequality \autoref{eq:11} we just established, we can define \textit{resisual entanglement} $\tau_{ABC}$ (though we haven't prove its invariance under permutation transformation):
\begin{align}
\tau_{ABC} \triangleq C^2_{A(BC)} - (C^2_{AB}+C^2_{BC}) = 2(\lambda^{AB}_1\lambda^{AB}_2 + \lambda^{AC}_1\lambda^{AC}_2) \ge 0
\end{align}
We first evaluate the product $\lambda^{AB}_1\lambda^{AB}_2$.
According to the discussion in the beginning of \autoref{sec:2.2.1}, 
if we consider the action of $\rho_{AB}\tilde\rho_{AB}$ restricted to its range, then $\lambda^{AB}_1\lambda^{AB}_2$ will be the 
square root of the determinant of this restricted transformation which we denote $R$, a $2\times2$ matrix.

Now we first give an explicit expression of $R_{ij}$.
Note that $span(\text{ran} (\rho_{AB})) = \{|\Phi_0\rangle,|\Phi_1\rangle\}$, 
where
$$
\begin{cases}
    |\Phi_0\rangle= a_{ij0}|ij\rangle \\ 
    |\Phi_1\rangle= a_{ij1}|ij\rangle \\ 
\end{cases}
$$
Then we consider how $\rho_{AB}\tilde\rho_{AB}$ will transform $\{|\Phi_0\rangle,|\Phi_1\rangle\}$, 
then re-arange it into matrix form then we can get $R_{ij}$:
\begin{align}
    R_{ij} = a_{rqj}a^\star_{mni} \epsilon_{mm'}\epsilon_{nn'} a^\star_{m'n'p}a_{r'q'p}\epsilon_{r'r}\epsilon_{q'q}
\end{align}
Then we give the expression of $\det R$ directly, for the algebra is "tedious but straightforward":
\begin{align}
\lambda^{AB}_1\lambda^{AB}_2 = \sqrt{\det R} = |d_1-2d_2+4d_3|
\end{align}
where
\begin{align*}
d_1 &= a^2_{000}a^2_{111}+a^2_{001}a^2_{110}+a^2_{010}a^2_{101}+a^2_{100}a^2_{011}\\
d_2 &= a_{000}a_{111}a_{011}a_{100} + a_{000}a_{111}a_{101}a_{010}\\
    &+ a_{000}a_{111}a_{110}a_{001}+ a_{011}a_{100}a_{101}a_{010}\\
    &+ a_{011}a_{100}a_{110}a_{001}+ a_{101}a_{010}a_{110}a_{001}\\
d_3 &= a_{000}a_{110}a_{101}a_{011}+a_{111}a_{001}a_{010}a_{100}
\end{align*}
The results look intimidating, but with extremely beautiful geometric picture.
We can visualize $a_{ijk}$ attached to the corner of a cube, 
and each term in $d_i$ is the product of four of the $a_{ijk}$ such that the "center of mass" of four is exactly the center of the cube.
And these terms can be exactly categorized into three class:
body diagonal and each used twice($d_1$), diagonal plane($d_2$), and two skew lines configuration($d_3$).

With this visualization, we can easily discover $\lambda^{AB}_1\lambda^{AB}_2$ is invariant under permutation transformation. 
Thus we easily reach our conclusion:
\begin{align}\label{eq:15}
\tau_{ABC} = C^2_{A(BC)} - (C^2_{AB}+C^2_{BC})= 4|d_1-2d_2+4d_3|
\end{align}
and the resisual entanglement $\tau_{ABC}$ is invariant under permutation transformation. 
The \autoref{eq:15} and the invariance are the main results of this paper.
\begin{figure}[h]\label{fig:1}
    \centering
    \includegraphics[scale=0.35]{pics/cube.png}
    \caption{Geometric Visualization of Terms in $d_i$}
\end{figure}

\section{Discussion and Conclusion}
\subsection{Generalization to Mixed State Situation}
Next we want to show our inequality could be generalized to situation where ABC is in mixed state. 

Now since ABC is in mixed state, we can no longer treat BC as a whole and define $C_{A(BC)}$ directly.
Let ABC in mixed state $\rho$, we have: 
$$
\langle C^2_{A(BC)} \rangle= \sum_i p_i C^2_{A(BC)}(\Phi_i)
$$
And define:
$$
C^2_{A(BC)}(\rho) = \inf \langle C^2_{A(BC)} \rangle
$$
Notice that the square of concurrence is what actually traded off in the inequality, 
therefore we propose our definition with the expectance of the square of concurrence. 

Average the inequality on the corresponding decomposition of $\inf \langle C^2_{A(BC)} \rangle$, 
and utilize the convexity of concurrence on the set of density operators\cite{mixed_state_entanglement}:
\begin{align*}
    LHS &= C^2_{AB}(\rho) + C^2_{AC}(\rho)\\
        &= C^2_{AB}(\sum_i p_i\Phi_i) + C^2_{AC}(\sum_i p_i\Phi_i)\\
        &\le \sum_i p_i (C^2_{AB}(\Phi_i) + C^2_{AC}(\Phi_i) ) \\ 
        &\le \sum_i p_i C^2_{A(BC)}(\Phi_i)\\
        &= C^2_{A(BC)}(\rho) = RHS
\end{align*}
Therefore, we prove that the inequality still holds for mixed state of ABC.

\subsection{Outlook}
It would be very interesting to know which of the results
of this paper generalize to larger objects or to larger collections
of objects.
It is not clear that whether we have a way to generalize the above 
results to systems having more energy levels, 
for the concurrence definition clearly requires some modification.
However, generalizing the conclusion to systems with more qubits seems to be more promising. 

Also, it is worth mentioning some other interesting perspectives and methodologies to research in this area.
In 1971, Roger Penrose suggested the idea of \textit{quantum tensor network} to represent tensor and help with tensor calculation.
In this paper\cite{tensor_network}, it gives an extremely clear and fast way to do the calculation and yields the correct results. 
And from this perspective, it could be easier to develop new ways to utilize quantum entanglement.

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\end{document}